Question { 1420 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.1 : (i) In the pricing of engineering bonds, 3 sets of data for Portfolio Value, Probability, Senior Tranche and Junior Tranche are : $2000, 81 %, $1000, $1000; $1000, 18 %, $1000, $0; $0, 1 %, $0, $0. By assuming independent defaults, find the price for : (a) Senior Tranche; (b) Junior Tranche. (ii) Assuming statistical independence of the values in the sample, the standard deviation of the mean (S) is related to the standard deviation of the distribution (s) by : N x S x S = s x s, where N is the number of observations in the sample used to estimate the mean. In a drug development project, let s = 1. Find the value of S if such a similar project is performed 100 times.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.1 : (i)(a) Price for Senior Tranche = 99 % x $1000 + 1 % x $0 = $990. (b) Price for Junior Tranche = 81 % x $1000 + 19 % x $0 = $810. (ii) Let N = 100, s = 1. Then S x S = s x s / N = 1 / 100, S = 1 / 10. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1504 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.2 : (a) In the pricing of a coupon bond, the formula is : P = c / (1 + r) + (c + B) / [ (1 + r) (1 + r) ] for 2 years to maturity, where c = annual coupon payment (in dollars, not a percent), B = par value, P = purchase price. Five years ago someone bought a 20 year coupon bond and would like to get rid of it now : A coupon rate of 7 %, it matures in exactly 2 years, par value is $1000, current interest rate is 5 %. (i) Find the value of c. (ii) Find the value of r or interest rate. (iii) Find the value of P. (iv) Guess the formula for P when the maturity period is 3 years, if such formula for 1 year duration is P = (c + B) / (1 + r). (b) In lemma of Ito on a Forward, recall that a forward contract is priced at : ln F = ln S + rT. Find the value of F in 5 decimal points when S = $100, r = interest rate = 0.05 / year, T = duration = 1 year.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.2 : (a) (i) c = $1000 x 0.07 = $70. (ii) r = 5 % = 0.05. (iii) P = c / (1 + r) + (c + B) / [ (1 + r) (1 + r) ] = 70 / (1 + 0.05) + (70 + 1000) / [ (1 + 0.05) (1 + 0.05) ] = $1037.19. (iv) For 3 year maturity period prediction by induction, P = c / (1 + r) + c / [ (1 + r) (1 + r) ] + (c + B) / [ (1 + r) (1 + r) (1 + r) ]. (b) Let ln F = ln S + rT = ln 100 + 0.05. Then F = $105.127109 or F = $105.12711 (in 5 decimal points). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1610 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.3 : (a) In the M / M / 1 queue that happens with randomness, let State 0 = the queue and server are empty, State 1 = the server is in use and the queue is empty, State 2 = the server is in use and 1 is in the queue, State 3 = the server is in use and 2 in the queue. Let P (0) = probability of State 0, P (1) = probability of State 1, P (2) = probability of State 2, P (3) = probability of State 3 and so on. If c = constant, P (1) = c P (0), P (2) = c [ c P (0) ], P (3) = c { c [ c P (0) ] }, write an equation that involves P (N), P (N + 1) and c. (b) Let L = market price of risk, r = riskless rate, m = expected return, s = volatility. Given that L = (m - r) / s related to oil prices, expected return = 12 %, s = 20 %, riskless rate = 8 %, calculate the market price of risk.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.3 : (a) P (1) = c P (0), P (2) = c [ c P (0) ] = c P (1), P (3) = c { c [ c P (0) ] } = c P (2). Then P (N + 1) = c P (N). (b) Let m = 12 %, r = 8 %, s = 20 %, then L = (m - r) / s = (12 - 8) / 20 = 4 / 20 = 1 / 5. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1249 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.4 : A university is enrolling new students of biochemical engineering degree. A long queue is formed during registration. Let L = rate of newcomers to a queue, m = number of clients served at a certain time, T = time in system. In M / M / 1 queue, let T = time waiting in a queue + service time, L = 2 / second, m = 3 / second. (a) Find the service time, A = 1 / m. (b) Calculate time waiting in a queue, B = A (L / m) / (1 - L / m). (c) What is the value of T?

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.4 : (a) A = 1 / m = 1 / 3 second. (b) B = A (L / m) / (1 - L / m) = (1 / 3) (2 / 3) / (1 - 2 / 3) = 2 / 3 second. (c) T = A + B = 1 / 3 + 2 / 3 = 1 second. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1387 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.5 : In an American style option for share market, 2 persons - A and B agree to the following : B is required to sell 100 shares of IBMS to A for $85 per share anytime that A wants in the next 8 months. A will pay B $2 per share up front, non-refundable for this option. IBMS involves in petrochemical processing. IBMS stock is currently selling for $80 per share. (a) If A did not buy the share of IBMS from B after 8 months, how much will B earn? (b) If the share of IBMS goes up to $100 / share in 6 months later : (i) how much should A pay B for 100 shares according to their optional agreement? (ii) how much will A earn from 100 shares purchased from B when all the 100 shares are sold to the open market? (iii) how much net profit will A earn for selling 100 shares to the open market?

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.5 : (a) Non-refundable payment = $2 / share x 100 shares = $200. (b) (i) Payment of A to B = $85 / share x 100 shares = $8500. (ii) Payment of open market to A = $100 / share x 100 shares = $10000. (iii) Net profit = answer b (ii) - answer b (i) - answer a = $10000 - $8500 - $200 = $1300. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1392 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.6 : In a random walk, a stochastic process starts off with a score of A. This is a score for a chemical engineering test. At each discrete event, there is probability p chance you will increase your score by B and a (1 - p) chance you will decrease your score by C. The event happens T times. Let D = expected value of score. (a) Form an equation of D as a function of A, B, C, T and p. (b) Find the value of D if A = 60, B = 1, C = -1, T = 50, p = 0.5.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.6 : (a) D = A + T [ pB + (1 - p) C ]. (b) D = A + T [ pB + (1 - p) C ] = 60 + 50 [ 0.5 (1) + (1 - 0.5) (-1) ] = 60. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1256 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.7 : An engineering investor purchases a $1000 bond that matures in 25 years. The coupon rate is 8 %. The purchase price is at 95 or 95 % of the par bond value. (a) What is the purchase price of the bond? (b) How much discount is enjoyed by the investor on average each year? (c) Find the income generated from coupon rate each year. (d) Find the overall interet rate or Yield to Maturity (YTM) by using the formula : YTM = [ Answer in (c) + Answer in (b) ] x 200 / [ par bond value + Answer in (a) ].

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.7 : (a) Purchase price = 0.95 x par bond value = 0.95 x $1000 = $950. (b) Annual discount = (total discount) / (25 years) = (0.05 x par bond value) / 25 = 0.05 x $1000 / 25 = $2. (c) Annual coupon rate payment = par value x coupon rate = $1000 x 0.08 = $80. (d) YTM = [ Answer in (c) + Answer in (b) ] x 200 / [ par bond value + Answer in (a) ] = (80 + 2) x 200 / (1000 + 950) = 8.41 %. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1378 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.8 : Cash flow agreement where the sign + or - of the amounts are start negative and at a point switch to positive and stay positive is referred to has an investment. The cash flow could be denoted by the symbol of [a, b, c ...]. Consider the following two cases in a biochemical factory : (a) You pay $1 today. You receive $1.05, one year from now. (b) You pay $3 today. You receive $5, one year from today. You pay $2, two years from today. Determine if cases (a) and (b) are an investment or a business transaction with cash flow symbol. Find the interest rate r in the cash flow in both cases (a) and (b).

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.8 : Case (a) : An investment [-1, 1.05]. Let 1.05 = 1 (1 + r), r = 0.05 or 5 %. Case (b) : A business transaction [-3, 5, -2]. Let 5 (1 + r) = 2 + 3 (1 + r) ^ 2, 3 (1 + r) ^ 2 - 5 (1 + r) + 2 = 0. Let s = 1 + r, then 3s ^ 2 - 5s + 2 = 0 = (3s - 2) (s - 1). Then s = 2 / 3 or 1. When s = 2 / 3, r = s - 1 = 2 / 3 - 1 = - 1 / 3 or - 33.33 %. When s = 1, r = s - 1 = 1 - 1 = 0 %. Let ^ be the symbol of power. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1286 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.9 : In the modelling of the total of n rolls of a dice by an engineering student, let D be the random outcome of rolling a dice once. (a) Find the probability of outcome of D = 1, 2, 3, 4, 5 and 6. (b) Find the average score of each rolling of a dice D. (c) Find the expected value, Sn of n rolls of a dice in term of n and D. A new dice has a value of D* = D - 3.5. (d) Find the values of D* for each volume of D = 1, 2, 3, 4, 5 and 6. (e) Find the equivalent model of Sn in term of n and D. (f) Find the expected value of D*.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.9 : (a) P (1) = P (2) = P (3) = P (4) = P (5) = P (6) = 1 / 6. (b) Average score of D = (1 + 2 + 3 + 4 + 5 + 6) / 6 = 3.5. (c) Expected value Sn = n (average score) = 3.5 n. (d) D* (D = 1) = 1 - 3.5 = - 2.5, D* (D = 2) = 2 - 3.5 = - 1.5, D* (D = 3) = 3 - 3.5 = - 0.5, D* (D = 4) = 4 - 3.5 = 0.5, D* (D = 5) = 5 - 3.5 = 1.5, D* (D = 6) = 6 - 3.5 = 2.5. (e) Sn = 3.5 n + nD*. (f) Expected value of D* = average of D* = [ D* (D = 1) + D* (D = 2) + D* (D = 3) + D* (D = 4) + D* (D = 5) + D* (D = 6) ] / 6 = (- 2.5 - 1.5 - 0.5 + 0.5 + 1.5 + 2.5) / 6 = 0. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1401 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.10 : Let D be the random outcome of rolling a dice once. A new dice has values of D* = D - 3.5. There is a total of n rolls of a dice. (a) Find the variance for D* by using the formula 6 V = [ D* (D = 1) ] [ D* (D = 1) ] + [ D* (D = 2) ] [ D* (D = 2) ] + [ D* (D = 3) ] [ D* (D = 3) ] + [ D* (D = 4) ] [ D* (D = 4) ] + [ D* (D = 5) ] [ D* (D = 5) ] + [ D* (D = 6) ] [ D* (D = 6) ]. (b) Calculate the standard deviation of D* as a square root of V. (c) Another new dice has values of D** = kD*. (i) Find the value of k so that D** has a standard deviation of 1. (ii) Find the values of D** for each outcome of D = 1, 2, 3, 4, 5 and 6, when the standard deviation is 1. (iii) Given that the average score of a dice is 3.5, find the equivalent, new and improved model of a dice, Sn in term of n and D**. (iv) Find the expected value of D** as the average of D**.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.10 : (a) 6 V = (1 - 3.5) (1 - 3.5) + (2 - 3.5) (2 - 3.5) + (3 - 3.5) (3 - 3.5) + (4 - 3.5) (4 - 3.5) + (5 - 3.5) (5 - 3.5) + (6 - 3.5) (6 - 3.5) = 17.5. V = 17.5 / 6 = 2.92. (b) Standard deviation = square root of V = (2.92) ^ (0.5) = 1.71. (c) (i) k = 1 / (standard deviation of D*) = 1 / 1.71 = 0.585. (ii) D** (D = 1) = -2.5 / 1.71 = -1.462, D** (D = 2) = -1.5 / 1.71 = -0.877, D** (D = 3) = -0.5 / 1.71 = -0.292, D** (D = 4) = 0.5 / 1.71 = 0.292, D** (D = 5) = 1.5 / 1.71 = 0.877, D** (D = 6) = 2.5 / 1.71 = 1.462. (iii) Sn = 3.5 n + 1.71 nD**. (iv) Expected value of D** = [ D** (D = 1) + D** (D = 2) + D** (D = 3) + D** (D = 4) + D** (D = 5) + D** (D = 6) ] / 6 = (- 1.462 - 0.877 - 0.292 + 0.292 + 0.877 + 1.462) / 6 = 0. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1451 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.11 : Let an Accounting Equation to be : Assets = Liabilities + Equity of Owner. In a biochemical engineering organization, its account balance sheet contains the following information : Accounts payable $2200; accounts receivable $1500; line of credit $500; cash $100; long term debt $3800; inventory $1300; investments $800; plant equipment $6800. Find the equity of owner by applying Accounting Equation.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.11 : Total asset = accounts receivable $1500 + cash $100 + inventory $1300 + investments $800 + plant equipment $6800 = $10500. Total liabilities = accounts payable $2200 + line of credit $500 + long term debt $3800 = $6500. Equity of owner = total assets - total liabilities = $10500 - $6500 = $4000. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1296 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.12 : You, as an engineer, buy a piece of biochemical instrument to start a servicing business. The instrument has a useful life of 3 years, that costs $60k. The instrumental business generates $100k per year. The labor costs $50k per year. The $60k new instrument depreciates in its value evenly up to 3 years, thereafter the old instrument needs to be replaced by another piece of new instrument. (i) In the first year of this engineering business : (a) what is its starting cash? (b) calculate the depreciation value of the instrument. (c) find the operating profit generated. (d) what is the capital expense of the business? (e) calculate the cash from operation. (f) calculate the ending cash. (ii) Based on the income statement of the business in the second year : (a) find the capital expense of the business. (b) what is its starting cash? (c) calculate the depreciation value of the instrument. (d) find the operating profit generated. (e) calculate the cash from operation. (f) calculate the ending cash.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.12 : (i)(a) Starting cash = $60k = cost of instrument. (b) Depreciation = instrumental (cost / useful life) = $60k / 3 years = $20k / year. (c) Operating profit = revenue - labor cost - depreciation = $100k - $50k - $20k = $30k. (d) Capital expense = $60k = cost of instrument. (e) Cash from operation = operating profit + depreciation = $30k + $20k = $50k. (f) Ending cash = starting cash + cash from operation - capital expense = $60k + $50k - $60k = $50k. (ii) (a) Capital expense of year 2 = 0. (b) Starting cash of year 2 = ending cash of year 1 = $50k. (c) Depreciation value = answer in (i)(b) = $20k / year. (d) Operating profit = answer in (i)(c) = $30k. (e) Cash from operation = answer in (i)(e) = $50k. (f) Ending cash = answer in (ii)(b) + answer in (ii)(e) - answer in (ii)(a) = $50k + $50k - 0 = $100k. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1333 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.13 : (i) In the Present Value Multiplication Rule, let PV = present value, Ra = interest rate for first discount, A = duration for first discount; Rc = interest rate for second discount, C = duration for second discount. Let PV = [ 1 / (1 + Ra) ^ A ] [ 1 / (1 + Rc) ^ C ] where ^ is the symbol of power : 3 ^ 2 = 3 x 3, 2 ^ 3 = 2 x 2 x 2. (a) For discounts involving 8 % / year for 3 years and 10 % / year for 9 years, find the value of PV. (b) If Re = interest rate for third discount, E = duration of third discount, form a mathematical equation of PV as a function of A, C, E, Ra, Rc, Re. Note : Discounts are available in the purchase of certain biochemical engineering instruments. (ii) Let R = nominal interest rate related to growth rate of money, r = real interest rate related to growth rate of purchase power. If I = inflation, where the unit of R, r and I is %, find the mathematical relatonship of r as a function of R and I.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.13 : (i) (a) Let Ra = 0.08, A = 3; Rc = 0.1, C = 9. Then PV = [ 1 / (1 + Ra) ^ A ] [ 1 / (1 + Rc) ^ C ] = [ 1 / (1 + 0.08) ^ 3 ] [ 1 / (1 + 0.1) ^ 9 ] = 0.33666. (b) PV = [ 1 / (1 + Ra) ^ A ] [ 1 / (1 + Rc) ^ C ] [ 1 / (1 + Re) ^ E ]. (ii) r = R - I. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1194 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.13 (CORRECTION) : (i) In the Present Value Multiplication Rule, let PV = present value, Ra = interest rate for first discount, A = duration for first discount; Rc = interest rate for second discount, C = duration for second discount. Let PV = [ 1 / (1 + Ra) ^ A ] [ 1 / (1 + Rc) ^ C ] where ^ is the symbol of power : 3 ^ 2 = 3 x 3, 2 ^ 3 = 2 x 2 x 2. (a) For discounts involving 8 % / year for 3 years and 10 % / year for 9 years, find the value of PV. (b) If Re = interest rate for third discount, E = duration of third discount, form a mathematical equation of PV as a function of A, C, E, Ra, Rc, Re. Note : Discounts are available in the purchase of certain biochemical engineering instruments. (ii) Let R = nominal interest rate related to growth rate of money, r = real interest rate related to growth rate of purchase power. If I = inflation, where the unit of R, r and I is %, find the mathematical relationship of r as a function of R and I.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.13 (CORRECTION) : (i) (a) Let Ra = 0.08, A = 3; Rc = 0.1, C = 9. Then PV = [ 1 / (1 + Ra) ^ A ] [ 1 / (1 + Rc) ^ C ] = [ 1 / (1 + 0.08) ^ 3 ] [ 1 / (1 + 0.1) ^ 9 ] = 0.33666. (b) PV = [ 1 / (1 + Ra) ^ A ] [ 1 / (1 + Rc) ^ C ] [ 1 / (1 + Re) ^ E ]. (ii) r = R - I. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1291 }

EXAMPLE 34.14 : For a formula of y = [ 1 - 1 / (1 + r) ^ n ] / r, let y = present value, r = interest rate / year, n = number of years to future value of 1 : (a) Find a simple mathematical relationship of y as a function of r when n = 1 million years. (b) What is the present value of $10000, if the annual discount rate is 10 %, forever?

Answer

(a) When n = 1 million years, it could be approximated as n = infinity. Then y = [ 1 - 1 / (1 + r) ^ n ] / r = [ 1 - 1 / (1 + r) ^ (infinity) ] / r = [ 1 - 1 / (infinity) ] / r = (1 - 0) / r = 1 / r. (b) For future value of $10000 and n = infinity forever, use the equation of y = 1 / r for future value of 1 and interest rate r. As r = 10 % = 0.1, then y = x / r = $10000 / 0.1 = $100000, where x = future value. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.